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(D) Let the bond energy of $A_2$ be $x$. Then,the bond energy of $AB$ is $x$ and the bond energy of $B_2$ is $0.5x$.The reaction for the formation of

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$400$

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(D) Let the bond energy of $A_2$ be $x$. Then,the bond energy of $AB$ is $x$ and the bond energy of $B_2$ is $0.5x$.The reaction for the formation of $AB$ is:$\frac{1}{2} A_2 + \frac{1}{2} B_2 	o AB$; $\Delta H_f = -100 \, kJ \, mol^{-1}$.The enthalpy of reaction is given by the sum of bond energies of reactants minus the sum of bond energies of products:$\Delta H = [\frac{1}{2} BE(A_2) + \frac{1}{2} BE(B_2)] - [BE(AB)]$Substituting the values:$-100 = [\frac{1}{2}(x) + \frac{1}{2}(0.5x)] - x$$-100 = 0.5x + 0.25x - x$$-100 = -0.25x$$x = \frac{100}{0.25} = 400 \, kJ \, mol^{-1}$.

$AB$,$A_2$ and $B_2$ are diatomic molecules. If the bond enthalpies of $A_2$,$AB$ and $B_2$ are in the ratio $1:1:0.5$ and enthalpy of formation of $AB$ from $A_2$ and $B_2$ is $-100 \, kJ \, mol^{-1}$,what is the bond energy of $A_2$ in $kJ \, mol^{-1}$?

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